What is the integral of-ln(x^2-x+1) ?

The integral of-ln(x^2-x+1) is -xln(x^2-x+1)+1/2 ln(x^2-x+1)-sqrt(3)arctan(1/(sqrt(3))(2x-1))+2x-1+C

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integral of-ln(x^2-x+1)

\int - ln (x^{2} - x + 1) d x

- x ln (x^{2} - x + 1) + \frac{1}{2} ln (x^{2} - x + 1) - 3 arctan (\frac{1}{3} (2 x - 1)) + 2 x - 1 + C

Solution steps

\int - ln (x^{2} - x + 1) d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int ln (x^{2} - x + 1) d x

Complete the square

x^{2} - x + 1 : (x - \frac{1}{2})^{2} + \frac{3}{4}

= - \int ln ((x - \frac{1}{2})^{2} + \frac{3}{4}) d x

Apply u-substitution

: \int ln (u^{2} + \frac{3}{4}) d u

= - \int ln (u^{2} + \frac{3}{4}) d u

Apply Integration By Parts

: u ln (u^{2} + \frac{3}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} + 3} d u

= - (u ln (u^{2} + \frac{3}{4}) - \int \frac{8 u ^{2}}{4 u ^{2} + 3} d u)

\int \frac{8 u ^{2}}{4 u ^{2} + 3} d u = - 3 arctan (\frac{2}{3} u) + 2 u

= - (u ln (u^{2} + \frac{3}{4}) - (- 3 arctan (\frac{2}{3} u) + 2 u))

Substitute back

u = x - \frac{1}{2}

= - ((x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} + \frac{3}{4}) - (- 3 arctan (\frac{2}{3} (x - \frac{1}{2})) + 2 (x - \frac{1}{2})))

Simplify

- ((x - \frac{1}{2}) ln ((x - \frac{1}{2})^{2} + \frac{3}{4}) - (- 3 arctan (\frac{2}{3} (x - \frac{1}{2})) + 2 (x - \frac{1}{2}))) : - x ln (x^{2} - x + 1) + \frac{1}{2} ln (x^{2} - x + 1) - 3 arctan (\frac{1}{3} (2 x - 1)) + 2 x - 1

= - x ln (x^{2} - x + 1) + \frac{1}{2} ln (x^{2} - x + 1) - 3 arctan (\frac{1}{3} (2 x - 1)) + 2 x - 1

Add a constant to the solution

= - x ln (x^{2} - x + 1) + \frac{1}{2} ln (x^{2} - x + 1) - 3 arctan (\frac{1}{3} (2 x - 1)) + 2 x - 1 + C

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What is the integral of-ln(x^2-x+1) ?
The integral of-ln(x^2-x+1) is -xln(x^2-x+1)+1/2 ln(x^2-x+1)-sqrt(3)arctan(1/(sqrt(3))(2x-1))+2x-1+C